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Chapter 7: Problem 1

Prove that every uniformly convergent sequence of bounded functions is uniformly bounded.

Short Answer

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Question: Prove that every uniformly converging sequence of bounded functions is uniformly bounded. Answer: To prove that every uniformly converging sequence of bounded functions is uniformly bounded, we first assume a sequence {f_n} of bounded functions that converges uniformly to a function f. Since each f_n is bounded, there exist constants M_n. Using the definition of uniform convergence and properties of absolute values and the triangle inequality, we find an upper bound for |f(x)|. We then determine the maximum of M_n and the bound for f(x) to show that the sequence is uniformly bounded. Thus, we have shown that every uniformly converging sequence of bounded functions is uniformly bounded.

Step by step solution

01

Begin the proof by assuming a uniformly converging sequence of bounded functions

Let {f_n} be a sequence of bounded functions that converges uniformly to a function f, where f_n : A 鈫� 鈩� for some non-empty set A and for every n 鈭� 鈩�. Since each f_n is a bounded function, there exists constants M_n 鈮� 0 such that for all x 鈭� A, we have |f_n(x)| 鈮� M_n.
02

Establish uniform convergence

Since {f_n} converges uniformly to f, for every 蔚 > 0, there exists a natural number N such that for all n 鈮� N and all x 鈭� A, |f_n(x) - f(x)| < 蔚. We can set 蔚 = 1 (as 蔚 can be any positive number) and find the corresponding N. Let's denote the subscript for this N as N蔚=1.
03

Find an upper bound for f(x) using the triangle inequality

Now, let's find an upper bound for |f(x)|. For x 鈭� A and n 鈮� N蔚=1, we have: |f(x)| = |f(x) - f_n(x) + f_n(x)| 鈮� |f(x) - f_n(x)| + |f_n(x)| By the definition of uniform convergence (with 蔚 = 1), we know that |f(x) - f_n(x)| < 1. And since f_n is a bounded function, |f_n(x)| 鈮� M_n. Therefore, |f(x)| 鈮� |f(x) - f_n(x)| + |f_n(x)| < 1 + M_n
04

Determine the maximum of M_n and the bound for f(x)

Since the inequality holds for all x 鈭� A and n 鈮� N蔚=1, we can find the maximum of 1 + M_n for all n 鈮� N蔚=1: M = max{1 + M_n | n 鈮� N蔚=1}
05

Complete the proof showing that the sequence is uniformly bounded

We now have the following inequality for all x 鈭� A, n 鈮� N蔚=1: |f(x)| < M Together with the initial bounds for |f_n(x)|, |f_n(x)| 鈮� M_n for n < N蔚=1. So, for all x 鈭� A and all n 鈭� 鈩�, we have: |f_n(x)| 鈮� max(M, M_1, M_2, ..., M_N蔚=1) This shows that the sequence {f_n} is uniformly bounded.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bounded Functions
A function is termed 'bounded' when its values do not exceed a certain fixed value, no matter the input values within the domain. This concept plays a crucial role in understanding uniformly convergent sequences and their properties.

For a function \( f_n: A \to \mathbb{R} \), being bounded means there exists a constant \( M_n \geq 0 \) such that for every \( x \in A \), we have \(|f_n(x)| \leq M_n \). Simply put, within the set \( A \), the function does not produce values larger than \( M_n \) in absolute terms.
  • A function is bounded over a set if its range over that set has both upper and lower limits.
  • This can also mean that the absolute value of the function does not exceed a particular value for any x in its domain.
Understanding this concept is key in mathematical analysis, particularly for proving other properties like uniform boundedness, where you verify these bounds apply uniformly across sequences.
Triangle Inequality
The triangle inequality is a fundamental concept in mathematics, especially in analysis and geometry. It suggests that for any real numbers or vectors, the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the remaining side.

In more formal terms, for real numbers \( a, b \) we have:
  • \(|a + b| \leq |a| + |b|\)
  • This inequality is pivotal when dealing with sequences and series, as it helps estimate bounds.
Within the context of sequences of functions, consider two functions \( f \) and \( f_n \). The inequality can be applied as:
\[|f(x)| \, = \, |f(x) - f_n(x) + f_n(x)| \, \leq \, |f(x) - f_n(x)| + |f_n(x)|\]This step enables us to establish bounds that aid in proving uniform boundedness of a sequence of functions and is often used as part of proof techniques in analysis.
Uniformly Bounded
When a sequence of functions \( \{f_n\} \) is uniformly bounded, it means there is a single bound that applies to every function in the sequence, uniformly across its domain. It simplifies dealing with sequences by assuring us that all function values in the sequence do not exceed a certain magnitude.

To say \( \{f_n\} \) is uniformly bounded means that there exists a constant \( M \) such that for every \( n \) and for all \( x \in A \), the condition \(|f_n(x)| < M\) holds. This property is stronger than each \( f_n \) being individually bounded鈥攈ere, the bounding constant \( M \) does not depend on \( n \) or \( x \), but rather applies to all simultaneously.
  • This concept is useful in evaluating the behavior of function sequences and in ensuring stability as functions converge.
  • Uniform boundedness is critical in proving convergence properties, as seen in the provided exercise.
Recognizing and using uniform bounds helps to substantiate the regular behavior of function sequences across their entire domain without having to assess each function individually.

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Most popular questions from this chapter

If $$ I(x)=\left\\{\begin{array}{ll} 0 & (x \leq 0) \\ 1 & (x>0) \end{array}\right. $$ if \(\left\\{x_{n}\right\\}\) is a sequence of distinct points of \((a, b)\), and if \(\Sigma\left|c_{n}\right|\) converges, prove that the series $$ f(x)=\sum_{n=1}^{\infty} c_{n} I\left(x-x_{n}\right) \quad(a \leq x \leq b) $$ converges uniformly, and that \(f\) is continuous for every \(x \neq x_{n}\).

Consider $$ f(x)=\sum_{n=1}^{\infty} \frac{1}{1+n^{2} x} $$ For what values of \(x\) does the series converge absolutely? On what intervals does it converge uniformly? On what intervals does it fail to converge uniformly? Is \(f\) continuous wherever the series converges? Is \(f\) bounded?

If \(f\) is continuous on \([0,1]\) and if $$ \int_{0}^{1} f(x) x^{n} d x=0 \quad(n=0,1,2, \ldots) $$ prove that \(f(x)=0\) on \([0,1]\). Hint: The integral of the product of \(f\) with any polynomial is zero. Use the Weierstrass theorem to show that \(\int_{0}^{1} f^{2}(x) d x=0\).

Let \(K\) be a compact metric space, let \(S\) be a subset of \(\mathcal{C}(K) .\) Prove that \(S\) is compact (with respect to the metric defined in Section \(7.14\) ) if and only if \(S\) is uniformly closed, pointwise bounded, and equicontinuous. (If \(S\) is not equicontinuous, then \(S\) contains a sequence which has no equicontinuous subsequence, hence has no subsequence that converges uniformly on \(K .\) )

Let \(f\) be a continuous real function on \(R^{1}\) with the following properties: \(0 \leq f(t) \leq 1, f(t+2)=f(t)\) for every \(t\), and $$ f(t)=\left\\{\begin{array}{ll} 0 & (0 \leq t \leq t) \\ 1 & (f \leq t \leq 1) . \end{array}\right. $$ Put \(\Phi(t)=(x(t), y(t))\), where $$ x(t)=\sum_{n=1}^{\infty} 2^{-n f}\left(3^{2 n-1} t\right), \quad y(t)=\sum_{n=1}^{\infty} 2^{-n} f\left(3^{2 n} t\right) . $$ Prove that \(\Phi\) is continuous and that \(\Phi\) maps \(I=[0,1]\) onto the unit square \(I^{2} \subset R^{2}\). If fact, show that \(\Phi\) maps the Cantor set onto \(I^{2}\). Hint: Each \(\left(x_{0}, y_{0}\right) \in I^{2}\) has the form $$ x_{0}=\sum_{n=1}^{\infty} 2^{-n} a_{2 n-1}, \quad y_{0}=\sum_{n=1}^{\infty} 2^{-n} a_{2 n} $$ where each \(a_{i}\) is 0 or 1 . If $$ t_{0}=\sum_{i=1}^{\infty} 3^{-1-1}\left(2 a_{i}\right) $$ show that \(f\left(3^{k} t_{0}\right)=a_{k}\), and hence that \(x\left(t_{0}\right)=x_{0}, y\left(t_{0}\right)=y_{0}\). (This simple example of a so-called "space-filling curve" is due to \(\mathrm{I}\). \(\mathrm{J}\). Schoenberg, Bull. A.M.S., vol. 44, 1938, pp. 519.)
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